Problem: You need to put your reindeer, Rudy, Ezekiel, Gloopin, Bloopin, and Prancer, in a single-file line to pull your sleigh. However, Bloopin and Rudy are fighting, so you have to keep them apart, or they won't fly. How many ways can you arrange your reindeer?
Answer: Forget about the reindeer that can't be together for a second, and let's try to figure out how many ways we can arrange the reindeer if we don't have to worry about that. We can build our line of reindeer one by one: there are $5$ slots, and we have $5$ different reindeer we can put in the first slot. Once we fill the first slot, we only have $4$ reindeer left, so we only have $4$ choices for the second slot. So far, there are $5 \cdot 4 = 20$ unique choices we can make. We can continue in this way for the third reindeer, then the fourth, and so on, until we reach the last slot, where we only have one reindeer left and so we can only make one choice. So, the total number of unique choices we could make to get to an arrangement of reindeer is $5\cdot4\cdot3\cdot2\cdot1 = 120.$ Another way of writing this is $5!$, or $5$ factorial. But we haven't thought about the two reindeer who can't be together yet. There are $120$ different arrangements of reindeer altogether, so we just need to subtract all the arrangements where Bloopin and Rudy are together. How many of these are there? We can count the number of arrangements where Bloopin and Rudy are together by treating them as one double-reindeer. Now we can use the same idea as before to come up with $4\cdot3\cdot2\cdot1 = 24$ different arrangements. But that's not quite right. Why? Because you can arrange the double-reindeer with Bloopin in front or with Rudy in front, and those are different arrangements! So the actual number of arrangements with Bloopin and Rudy together is $24 \cdot 2 = 48$ So, subtracting the number of arrangements where Bloopin and Rudy are together from the total number of arrangements, we get $72$ arrangements of reindeer where they will fly.